A problem by Lorenc Bushi

Notice that:

$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2((ab)^2+(bc)^2+(ac)^2)$ .Since we already know that $a^2+b^2+c^2=4$ our problem becomes in finding the value of $(ab)^2+(bc)^2+(ac)^2$ .Observe that

$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=4$ .Since we know that $a+b+c=0$ easily we find $ab+bc+ac=-2$ .But the problem is not solved yet.We must find $(ab)^2+(bc)^2+(ac)^2$ .We can do the same thing to find our desired sum.

$(ab+bc+ac)^2-2abc(a+b+c)=(ab)^2+(bc)^2+(ac)^2$ .Since $a+b+c=0$ the whole term that involves $abc$ is canceled and then it is not hard for us to find $(ab)^2+(bc)^2+(ac)^2=4$ .Substituting this in the equation above we get our desired result which is $\boxed{8}$ .

We shall use here the identity $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$

Given that, $a+b+c=0$ ....(i) and $a^{2}+b^{2}+c^{2}=4$ .....(ii)

Squaring eq. (i), we get----

$(a+b+c)^{2}=0^{2}$

$\implies (a^{2}+b^{2}+c^{2})+2(ab+bc+ca)=0$

$\implies 4+2(ab+bc+ca)=0$ [From (ii)]

$\implies 2(ab+bc+ca)=-4$

$\implies ab+bc+ca=-2$

Squaring both sides again, we get---

$\implies ((ab)^{2}+(bc)^{2}+(ca)^{2})+2(ab^{2}c+bc^{2}a+ca^{2}b)=4$

$\implies ((ab)^{2}+(bc)^{2}+(ca)^{2})+2abc(b+c+a)=4$

$\implies ((ab)^{2}+(bc)^{2}+(ca)^{2})=4$ [From (i)]

$\implies a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}=4$ ......(iii)

Now, squaring eq. (ii), we get-----

$(a^{2}+b^{2}+c^{2})^{2}=4^{2}$

$a^{4}+b^{4}+c^{4}+2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})=16$

$a^{4}+b^{4}+c^{4}+2\times 4=16$ [From (iii)]

$a^{4}+b^{4}+c^{4}+8=16 \implies a^{4}+b^{4}+c^{4}=\boxed{8}$